Basic strategies (perhaps trivial but may be useful to someone). Fifth rule: Addition of corners, border, inner area. And why the numbers in the corners of a 4x4 must be the same as those in the inner area (this property may be used in the timed puzzles).

Let’s call “b” the sum of all numbers in the border (the “addition value” of the border). Let’s call “C” the sum of the four corners. Let’s call “I” the addition of all numbers in the inner area (the “value” of the inner area). Let’s call “p” the size of the puzzle and let’s call “s” the sum of the numbers of a line (row or column) (as we know this sum is 10, 15, 21, 28, 36, …, for puzzles 4x4, 5x5, 6x6, 7x7, 8x8, …).

If we sum all the numbers in the top and bottom rows and in the leftmost and rightmost columns, that total will be 4s (four times a line) and it will be equal to the “border” plus the four corners since we have counted the corners twice (the four corners are duplicated in this count). So:

b + C = 4s b + I = ps (this second equation because the border plus the inner area constitute the full puzzle and the sum of all numbers in the puzzle equals p multiplied by s (the “value” of a line). Subtracting these two equations:

I - C = (p - 4)s I = C + (p - 4)s or also C = I - (p - 4)s. Sometimes, in a 6x6 (p = 6, s= 21), the corners are given so: I = C + (6 – 4)s = C + 2s = C + 42 (or C = I - 42). When the corners are given, the inner area may be deduced easily, but in other cases, not so intuitive, the inner area is given (or easily calculated) and then we deduce C (if in this case one of the corners is known, for instance, we determine the sum of the other three and then we can analyze the combinations for the “virtual triangle” they constitute).

Now let’s see the special case of the 4x4’s: I = C + (4 - 4)s = C + 0 = C, the sum of the corners must be equal to the sum of the numbers in the inner area (the square 2x2 inside). But not only this. Further, the numbers in the corners and in the inner area must be the same. The demonstration is easy. Suppose that we have two equal numbers inside this “heart”, for instance, two 1’s. As these 1’s occupy (“erase”) columns B and C and rows 2 and 3, the other two 1’s must go to two (opposite) corners. Let’s see all the possibilities:

Nrs. inside the “heart” Corners

1123 11 23 (the rest 5 [for a total of 7] is only made with the 2-3, 1-4 producing three 1’s in the corners)1124 11 24 (only combination for a sum of 6, 3-3 is not possible because 3-3 in the other two corners force two 3’s inside the “heart” against the hypothesis)1134 11 34 (only combination for a sum of 7)2213 22 13 (obviously four 2’s in the corners are not possible)2214 22 14 (a sum of 5 with 2-3 is not possible, three 2’s in the corners)2234 22 34 (only combination for a sum of 7)3312 33 123314 33 14 (a sum of 5 with 2-3 impossible, three 3’s in the corners)3324 33 24 (obviously four 3’s in the corners are not possible)4412 44 124413 44 13 (2-2 is not possible since the other two 2’s would go the “heart” against the hypothesis)4423 44 23 (1-4 makes three 4’s in the corners).

Now, three equal numbers inside the “heart” are impossible (two equal numbers contiguous). Finally, if the four numbers inside the inner area are different, 1-2-3-4 (I = 10), the four numbers in the corners must also be 1-2-3-4 (C = 10), different numbers, because if two of them were equal, that would force the repeated number to go twice to the “heart” against the hypothesis.

An easy example (practical): The June, 20 4x4 difficult (Puzzle Id: 299530).The 3 in D1 forces a 3 in the “heart” but in B2 is not possible (it would require two consecutive 1’s to complete the cage “1-“); in C2 is not possible because 3 is not a factor of 8; in C3 is not possible because the 1 in A4 makes impossible a sum of 3 in C4-D4, so a 3 must go to B3 being the other 3 [prime factor of the “72x”] necessarily in A2; the rest is very quick (4 in B4, 2 in A3, etc.).

Looking to the general expression: I = C + (p-4)s, we observe that if “C” increases “I” also increases and vice versa; obviously if both increase simultaneously the “walls” in the border must decrease and vice versa. Let’s call “w” the sum of all numbers in the four walls (the border without the cornes):

Ok, maybe someone can follow this line of thinking. As for me, here's my explanation of why you get the same numbers in the inner area as in the corners.

We only need to explain why any one corner number is found in the inner area and it will be the same for all the other corners.

Let's assume the number 1 is in the top-left corner. x is for any number that is not 1.1xxxx???x???x???

Then the other three 1s have to be found in the ? area. It is impossible to not have a "1" in the inner area because you can only have at most one "1" in the right column and one "1" in the bottom row. The last "1" has to be in the inner area.

Since this is true for any corner number, you will find all the corner numbers in the inner area.

picklepep

Posted on:Wed Jul 06, 2011 4:20 am

Posts: 98Joined: Fri May 13, 2011 12:48 am

Re: Why in a 4x4 numbers in corners are those in inner area

This is very interesting, but I am not sure that it is generally useful. I think that most solutions can be worked out with quicker more straightforward methods.

sneaklyfox

Posted on:Wed Jul 06, 2011 4:27 am

Posts: 422Location: CanadaJoined: Fri May 13, 2011 2:43 am

Re: Why in a 4x4 numbers in corners are those in inner area

picklepep wrote:

This is very interesting, but I am not sure that it is generally useful. I think that most solutions can be worked out with quicker more straightforward methods.

This is very interesting, but I am not sure that it is generally useful. I think that most solutions can be worked out with quicker more straightforward methods.

I would agree with you.

The corners and the border in the 5x5’s and 6x6’s (answer to Sneaklyfox and Picklepep and a practical example of some utility of this rule). The Sneaklyfox demo for a 4x4 is abridged and concise. However all that paraphernalia I used was because I was really interested in doing some systematic analysis (mainly oriented to beginners). The idea is: First deriving the rule, second obtaining the corollary and practical use, if any. Let’s see the case of a 5x5. If we consider the equation of my first article: I = C + (p - 4)s, when applied to a 5x5, we get: I = C + (5 - 4) x 15 = C + 15. But in the core of a 5x5 we have 9 cells, why not deducing then intuitively that the corners and each of the five numbers 1, 2, 3, 4 and 5 (that have a sum of 15) could be inside?. Certainly that is what really happens: In every 5x5, the inner area (the core 3x3) is made by the four corners and once the numbers 1 thru 5. Lets see an example (the 5x5 difficult, July 05, 2011, Puzzle id: 314715):

Consider one of the numbers, i. e., the 3. Now we can affirm that, since it is impossible to have five 3’s in the border (only four lines) then one 3 must be necessarily inside (and the same happens with the other numbers 1, 2, 4 and 5). Now we have four idle cells in the “heart”. Let’s see the corners (following the Sneaklyfox outline): If a number “n” is in a corner, an additional “n” must go to the core (three in total to the moment) (because in the L-shaped “z” we can only locate the other two “n’s”, that makes the total of five n’s). And this is valid for any of the other three corners (see for instance the location of the two’s in the example, etc. (I use green for the corners, and blue for the numbers 1 thru 5 in the “heart”).

nxxxx x???z x???z x???z xzzzz

Now let’s apply the property to a 6x6. The equation gives: I = C + (p - 4)s = C + (6 - 4) x 21 = C + 42. Observing the equation we could initially estimate that, being 21 the sum of all number in a line, perhaps we would have in the inner area (16 cells), the 4 corners plus twice the six numbers 1, 2, 3, 4, 5 and 6, and this is what really happens in every 6x6. The demonstration is exactly the same: Any number can not be more than 4 times in the border so it must be twice in the inner area (12 numbers occupied) and for a corner “n” since we have 3 n’s already located (the corner and two more inside) we must locate an additonal “n” in the inner area since no more than two (among the pending three) can be situated in the L-shaped “z’s”. See an example (the 6x6 difficult, June 14, 2011, Puzzle id: 305714):

(I use the green again for the corners and blue and purple to show the two series of 1 thru 6). The rule of course can be extended to any size. What really happens in a 4x4 is that there is no “additional space” in the core (like a mathematical point of dimension zero) and then the corners go just once to the inner area, so the case of the 4x4 looks trivial, but it is curious to observe in general how the core grows when the puzzle grows. I agree that, in a 4x4, in very few cases, the rule is practical (the example I used in my first topic was an exception, in fact the number of all possible 4x4’s, excluding symmetries, is not too big, it is mainly a calculation with "sudokus", neither the number of different operations applied to them) and one in general will use other means. But now I am going to apply the property to a practical case in a 6x6 (the 6x6 difficult, July 07, 2011, Puzzle id: 318137).

Solution: First, of course, we place the “144x”, that is, 4-6-6, 4 in B2, 6 in C2 and 6 in B3. Now C = 1 + 6 + 4 + 3 = 14, so I = 14 + 42 = 56 (even number) (we may say that the “addition” of the border can be calculated directly and then obtain I subtracting from 126, the full puzzle, but this way is faster). Applying the parity rule the cage “2: “ is an odd cage, so the operands are 1-2-4 (sum of 7) (since 1-1-2 and 1-3-6 are even). The 4 must go to E3. Now the cage “0-“ has a sum of 6 ( to the total of 56 in the inner area) so the combination is 1-2-3. The cage “14+” has three possibilities 6-6-2 (not valid since only three 6’s are in the core as we have seen; here we may object : ”Well, we can see that because otherwise no 6’s could be placed in row 6"), the 5-5-4 is not possible due to the 4 in E3, so the only valid combinaron is 6-5-3. Only two 2’s (2 is not in a corner) can be inside so, since we already have them, “5+” is made with 1-4 and “8+” is made with 3-5 (not 2-6; also because another 5 must be in the core and the “8+” is the only possible place; once again we can argue: "Well, four 6’s are not possible in core due to row 6") (vice versa, if “8+” is 3-5, “5+” can not contain a fourth 3 “due to row 1” and due to the presence in core of two 2’s an three 3’s according to our exposition). Now we place the 6 of the “14+” in D5.

Now we go with the numbers in red: 6 in E6 and A4, 4 in D1, 1 in E2, 2 in E1, 2 in D2, 3 in A2 and 5 in FE2, 3 in D3 (see the 3’s in A2 and F6), 5 in D4, 3 in E4, 5 in E5 and 1 in D6. Now we situate the purple numbers of rows 3, 4 and 5 consecutively and quickly. The rest is trivial.

Conclusion: Of course there are many other ways of arriving to the solutions, but all possible tools may be used or can aid.

This is the “official” solution:

sneaklyfox

Posted on:Fri Jul 08, 2011 4:36 pm

Posts: 422Location: CanadaJoined: Fri May 13, 2011 2:43 am

Re: Why in a 4x4 numbers in corners are those in inner area

clm wrote:

The Sneaklyfox demo for a 4x4 is abridged and concise. However all that paraphernalia I used was because I was really interested in doing some systematic analysis (mainly oriented to beginners).

I suppose you could consider it an abridged version. But I more consider it the easy short method of saying the same thing. Like, if I wanted to arrive at a location 1 block North, I wouldn't go 1 block East, 1 block North, 1 block West. I would just go 1 block North. It just seemed unnecessary. But ok, you wanted it very mathematical in the terms you've used. That's fine. I don't really think it's oriented to beginners though. I think beginners would be satisfied to either just believe the premise or read a brief "proof".